Mock Test 5: MCQs with Explanations
1. Preparation of alkyl halides in laboratory is least preferred by
- (a) halide exchange
- (b) treatment of alcohols
- (c) addition of hydrogen halides to alkanes
- (d) direct halogenation of alkanes
Correct Answer: (d)
Explanation: Direct halogenation of alkanes is generally avoided in the laboratory because it often results in a complex mixture of mono-, di-, and poly-substituted products which are difficult to separate.
2. The main product of the given reaction would be: 2-butene + chloroform $\xrightarrow{NaOH}$ [Product] $\xrightarrow{Hydrolysis}$ ?
- (a) butanoic acid
- (b) 2-methylbutanoic acid
- (c) 1,1,1-trichloro-2-methylbutane
- (d) 1,4-butanediol
Correct Answer: (b)
Explanation: The reaction of an alkene with chloroform and NaOH generates a dichlorocarbene intermediate which adds across the double bond. Subsequent hydrolysis and rearrangement (similar to a Reimer-Tiemann or specific carbonylation logic) leads to the formation of 2-methylbutanoic acid.
3. Identify the set of reagents X and Y in: $CH_{3}CH_{2}CH_{2}Br \xrightarrow{X}$ Product $\xrightarrow{Y}$ $CH_{3}CH(Br)CH_{3}$
- (a) $X =$ dil. aq. NaOH; $Y = HBr/acetic acid$
- (b) $X =$ conc. alc. NaOH, $80^{\circ}C$; $Y = HBr/acetic acid, 20^{\circ}C$
- (c) $X =$ dil. aq. NaOH; $Y = Br_{2}/CHCl_{3}$
- (d) $X =$ conc. alc. NaOH; $Y = Br_{2}/CHCl_{3}$
Correct Answer: (b)
Explanation: Concentrated alcoholic NaOH at high temperature facilitates elimination (dehydrohalogenation) of propyl bromide to propene ($X$). Addition of HBr in acetic acid to propene then follows Markovnikov's rule to yield isopropyl bromide ($Y$).
4. Predict the correct stereoisomeric product for the addition of $Br_{2}$ to cis-2-butene.
- (a) d-form
- (b) l-form
- (c) racemic mixture
- (d) meso form
Correct Answer: (c)
Explanation: The addition of bromine to an alkene is an anti-addition process. When $Br_{2}$ adds to a cis-alkene, the resulting vicinal dihalide is formed as a racemic mixture (a pair of enantiomers).
5. o-Chlorotoluene can undergo: (i) electrophilic aromatic substitution, (ii) nucleophilic aromatic substitution, (iii) nucleophilic aliphatic substitution, (iv) free radical substitution.
- (a) only (i)
- (b) (i) and (iv)
- (c) (i), (ii) and (iv)
- (d) all four
Correct Answer: (c)
Explanation: o-Chlorotoluene can undergo electrophilic substitution on the ring (directed by $-CH_{3}$ and $-Cl$), nucleophilic aromatic substitution under drastic conditions (Dow's process logic), and free radical substitution at the methyl group ($CH_{3}$) in the presence of light/UV.
6. Regarding ethers: (1) Oxygen atom is $sp^{3}$-hybridised, (2) They are liquids at room temperature, (3) They are miscible with water, (4) They are very active. Correct statements are:
- (a) only 1
- (b) only 3 and 4
- (c) only 1 and 2
- (d) 1, 2 and 3
Correct Answer: (d)
Explanation: In ethers, the oxygen atom is $sp^{3}$ hybridised. Most common ethers are volatile liquids. Lower ethers show significant solubility (miscibility) in water due to hydrogen bonding. However, ethers are generally chemically inert (not "very active").
7. Which of the following statements is false?
- (a) Artificial silk is derived from cellulose.
- (b) Nylon-6,6 is an example of elastomer.
- (c) The repeat unit in natural rubber is isoprene.
- (d) Both starch and cellulose are polymers of glucose.
Correct Answer: (b)
Explanation: Nylon-6,6 is a fibre, characterized by strong intermolecular hydrogen bonding, not an elastomer. Natural rubber is an elastomer.
8. The metals present in insulin, haemoglobin and vitamin $B_{12}$ are respectively
- (a) Zn, Hg, Cr
- (b) Co, Fe, Zn
- (c) Mg, Fe, Co
- (d) Zn, Fe, Co
Correct Answer: (d)
Explanation: Insulin contains Zinc, Haemoglobin contains Iron, and Vitamin $B_{12}$ (cyanocobalamin) contains Cobalt.
9. Synthesis of each molecule of glucose in photosynthesis involves
- (a) 6 molecules of ATP
- (b) 18 molecules of ATP
- (c) 10 molecules of ATP
- (d) 8 molecules of ATP
Correct Answer: (b)
Explanation: In the Calvin cycle of photosynthesis, the synthesis of one molecule of glucose requires 18 molecules of ATP and 12 molecules of NADPH [This is standard biochemistry found in biology/chemistry synthesis chapters, though not explicitly detailed in the chemistry snippets].
10. When calcium acetate and calcium formate together are subjected to dry distillation, the product is
- (a) acetaldehyde
- (b) acetone
- (c) formaldehyde
- (d) none of these.
Correct Answer: (a)
Explanation: Dry distillation of a mixture of calcium acetate and calcium formate yields acetaldehyde ($CH_{3}CHO$).
11. Relative acidity of the following is in the order
- (a) $RCOOH > H_{2}CO_{3} > C_{6}H_{5}OH > H_{2}O > ROH$
- (b) $RCOOH > ROH > H_{2}CO_{3} > C_{6}H_{5}OH > H_{2}O$
- (c) $ROH > RCOOH > H_{2}CO_{3} > C_{6}H_{5}OH > H_{2}O$
- (d) $RCOOH > C_{6}H_{5}OH > ROH > H_{2}CO_{3} > H_{2}O$
Correct Answer: (a)
Explanation: Carboxylic acids are the strongest, followed by carbonic acid, then phenol, then water, and finally alcohols.
12. Compound $P (C_{5}H_{10}O)$ forms phenyl hydrazone, gives a negative Tollen's and iodoform test. $P$ on reduction gives n-pentane. $P$ is
- (a) [structure a] pentanal
- (b) [structure b] 2-pentanone
- (c) [structure c] 3-pentanone
- (d) none
Correct Answer: (c)
Explanation: Being $C_{5}H_{10}O$, it's an aldehyde or ketone. Negative Tollen's test indicates it's a ketone. Negative iodoform test means it is not a methyl ketone (so not 2-pentanone). Reduction to n-pentane confirms it's a straight-chain 5-carbon ketone. Thus, it must be 3-pentanone.
13. Which of the following compounds can be classified as aryl halides?
- (a) $p-ClC_{6}H_{4}CH_{2}CH(CH_{3})_{2}$
- (b) $p-CH_{3}CHCl(C_{6}H_{4})CH_{2}CH_{3}$
- (c) $o-BrH_{2}C(C_{6}H_{4})CH(CH_{3})CH_{2}CH_{3}$
- (d) $C_{6}H_{5}CH_{2}Cl$
Correct Answer: (a)
Explanation: An aryl halide is a compound where the halogen atom is directly bonded to an $sp^{2}$ carbon of an aromatic ring. In (a), Chlorine is directly on the benzene ring. In the others, the halogen is on a side-chain (aralkyl halides).
14. In the reaction, $C_{6}H_{5}CH_{2}Br \xrightarrow{(i)Mg, ether; (ii)H_{3}O^{+}}$ [Product], the product 'X' is
- (a) $C_{6}H_{5}CH_{2}OCH_{2}C_{6}H_{5}$
- (b) $C_{6}H_{5}CH_{2}OH$
- (c) $C_{6}H_{5}CH_{3}$
- (d) $C_{6}H_{5}CH_{2}CH_{2}C_{6}H_{5}$
Correct Answer: (c)
Explanation: Benzyl bromide reacts with Mg in ether to form the Grignard reagent $C_{6}H_{5}CH_{2}MgBr$. Acidic hydrolysis of this reagent replaces the $-MgBr$ group with a Hydrogen atom, yielding toluene ($C_{6}H_{5}CH_{3}$).
15. The conversion of ethanol to propane nitrile $(CH_{3}CH_{2}CN)$ is best carried out by
- (a) $CH_{3}CH_{2}CH_{2}OH + KCN \rightarrow$
- (b) $CH_{3}CH_{2}OH + HCN \rightarrow$
- (c) $CH_{3}CH_{2}OH \xrightarrow{TsCl/Pyr.} CH_{3}CH_{2}OTs \xrightarrow{KCN}$
- (d) $CH_{3}CH_{2}OH + CH_{3}CN \rightarrow$
Correct Answer: (c)
Explanation: Since $-OH$ is a poor leaving group, it is first converted into a good leaving group like a tosylate ($OTs$). Subsequent nucleophilic substitution with $KCN$ successfully introduces the nitrile group.
16. Which will undergo Friedel-Crafts alkylation reaction? (1) Toluene, (2) Ethylbenzene, (3) Benzoic acid, (4) Nitrobenzene.
- (a) 1, 2 and 4
- (b) 1 and 3
- (c) 1 and 2
- (d) 1 and 2
Correct Answer: (c)
Explanation: Friedel-Crafts alkylation requires an electron-rich or moderately deactivated aromatic ring. Toluene and Ethylbenzene (activated rings) undergo the reaction easily. Strongly deactivated rings like Benzoic acid and Nitrobenzene do not undergo Friedel-Crafts reactions.
17. Which of the following is an example of $S_{N}2$ reaction?
- (a) $CH_{3}Br + OH^{-} \rightarrow CH_{3}OH + Br^{-}$
- (b) $CH_{3}CH(Br)CH_{3} + OH^{-} \rightarrow CH_{3}CH(OH)CH_{3} + Br^{-}$
- (c) $CH_{3}CH_{2}OH \rightarrow CH_{2}=CH_{2}$
- (d) $(CH_{3})_{3}C-Br + OH^{-} \rightarrow (CH_{3})_{3}C-OH + Br^{-}$
Correct Answer: (a)
Explanation: $S_{N}2$ reactions are most favoured by primary alkyl halides (like methyl bromide) due to minimal steric hindrance. Tertiary halides (d) proceed via $S_{N}1$.
18. Which of the following reactions will not give N, N-dimethylbenzamide?
- (a) Benzoyl chloride + Dimethylamine
- (b) Benzonitrile + Methylmagnesium iodide
- (c) Benzoic acid + Dimethylamine
- (d) Benzoic anhydride + Dimethylamine
Correct Answer: (b)
Explanation: Reaction of a nitrile with a Grignard reagent followed by hydrolysis yields a ketone (acetophenone in this case), not an amide. Acyl halides (a), carboxylic acids (c), and anhydrides (d) can all react with amines to form amides.
19. Find the incorrect statement.
- (a) In aq. solution, only a very little fraction of glucose exists as open chain aldose.
- (b) Galactose is the epimer of glucose.
- (c) Sucrose is a leavorotatory sugar.
- (d) Fructose gives positive Fehling's test.
Correct Answer: (c)
Explanation: Sucrose is dextrorotatory (+). It is only after hydrolysis that the resulting mixture (invert sugar) becomes levorotatory. Fructose, though a ketone, gives a positive Fehling's test due to rearrangement in basic medium.
20. A $\beta$-pleated sheet
- (a) has C=O and N-H bonds in different planes.
- (b) -R groups are oriented above the plane of the sheet only.
- (c) can have antiparallel as well as parallel arrangement.
- (d) shows hydrogen bonding between alternate amino acid residues.
Correct Answer: (c)
Explanation: The $\beta$-pleated sheet secondary structure of proteins consists of polypeptide chains running side by side, which can be arranged in either a parallel or antiparallel fashion [This is standard biochemistry/polymers knowledge].
21. A compound A ($C_{7}H_{7}NO$) + $Br_{2}/KOH \rightarrow$ B (amine, carbylamine +ve). B diazotisation/coupling $\rightarrow$ azo dye. A is
- (a) $C_{6}H_{5}CONHCOCH_{3}$
- (b) $C_{6}H_{5}CONH_{2}$
- (c) $o, m$ or $p-C_{6}H_{4}(NH_{2})CHO$
- (d) $C_{6}H_{5}NO_{2}$
Correct Answer: (b)
Explanation: Compound A ($C_{7}H_{7}NO$) reacts with $Br_{2}/KOH$ (Hofmann Bromamide reaction) to give a primary amine B. This indicates A is an amide. Benzamide ($C_{6}H_{5}CONH_{2}$) yields aniline (B), which is a primary aromatic amine that gives the carbylamine test and forms azo dyes.
22. What form of glutamic acid would predominate in a strongly basic solution?
- (a) Zwitterionic form
- (b) Di-anionic form
- (c) Mono-anionic form
- (d) Cationic form
Correct Answer: (b)
Explanation: Glutamic acid is an acidic amino acid with two carboxyl groups. In a strongly basic solution, both carboxyl groups lose protons to form the di-anionic species.
23. (A) $\xrightarrow{reduction}$ (B) $\xrightarrow{CHCl_{3}/NaOH}$ (C) $\xrightarrow{reduction}$ N-methylaniline. (A) is
- (a) Nitrobenzene
- (b) Aniline
- (c) Benzonitrile
- (d) Benzamide
Correct Answer: (a)
Explanation: Nitrobenzene (A) is reduced to aniline (B). Aniline reacts with $CHCl_{3}/NaOH$ (carbylamine reaction) to form phenyl isocyanide (C). Reduction of phenyl isocyanide yields N-methylaniline.
24. Order of decreasing acidic character: cyclohexanol (I), acetic acid (II), 2,4,6-trinitrophenol (III), phenol (IV).
- (a) III > II > IV > I
- (b) II > III > IV > I
- (c) II > III > IV > I
- (d) III > IV > II > I
Correct Answer: (a)
Explanation: 2,4,6-Trinitrophenol (Picric acid) is exceptionally acidic due to three strong electron-withdrawing groups. Acetic acid is a stronger acid than unsubstituted phenol. Cyclohexanol is the least acidic.
25. Toluene $\xrightarrow{Nitration}$ Product $\xrightarrow{Sn/HCl}$ Product $\xrightarrow{Diazotisation}$ Product $\xrightarrow{CuBr/HBr}$ [Final Product].
- (a) o- and m-bromotoluenes
- (b) o- and p-bromotoluenes
- (c) o- and p-dibromobenzenes
- (d) o- and p-bromoanilines
Correct Answer: (b)
Explanation: Nitration of toluene gives a mixture of ortho- and para-nitrotoluenes (methyl is o/p directing). Reduction gives o- and p-toluidines. Diazotisation followed by Sandmeyer reaction with $CuBr$ replaces the amino groups with Bromine, yielding o- and p-bromotoluenes.
26. In the reaction, A $\xrightarrow{KCN}$ B $\xrightarrow{[H]}$ $C_{2}H_{5}NH_{2}$
- (a) A is $CH_{3}I$
- (b) B is $CH_{3}NC$
- (c) A is $C_{2}H_{5}I$
- (d) B is $C_{2}H_{5}NC$
Correct Answer: (a)
Explanation: Reduction of a nitrile yields a primary amine with the same number of carbons. For ethylamine ($C_{2}$), the nitrile must be acetonitrile ($CH_{3}CN$, $C_{2}$). Thus, A must be methyl iodide ($CH_{3}I$) which reacts with $KCN$ to form acetonitrile (B).
27. Phenol can be prepared by the reaction between
- (a) aniline and $HNO_{3}$ at 373 K
- (b) $C_{6}H_{5}MgBr$ and $CO_{2}$ followed by hydrolysis
- (c) $C_{6}H_{5}Cl$ and NaOH at 373 K
- (d) $C_{6}H_{5}SO_{3}Na$ and NaOH at 573-623 K
Correct Answer: (d)
Explanation: Fusion of sodium benzenesulphonate with molten NaOH at high temperatures (573-623 K) followed by acidification is a standard industrial method to prepare phenol.
28. If one strand of DNA is ATGCTTGA, the complementary strand is
- (a) TACGAACT
- (b) TCCGAACT
- (c) TACGTACT
- (d) TACGTAGT
Correct Answer: (a)
Explanation: In DNA, base pairing is specific: Adenine (A) pairs with Thymine (T), and Guanine (G) pairs with Cytosine (C). Thus, the complement to ATGCTTGA is TACGAACT.
29. Which of the following is not a protein test?
- (a) Biuret test
- (b) Millon's test
- (c) Lipoprotein test
- (d) Sakaguchi test
Correct Answer: (c)
Explanation: Biuret, Millon's, and Sakaguchi tests are standard chemical tests used to identify proteins or specific amino acids within them. Lipoprotein test is not a general protein identification test.
30. Phenol is
- (a) a base weaker than $NH_{3}$
- (b) an acid stronger than carbonic acid
- (c) an acid weaker than carbonic acid
- (d) neutral
Correct Answer: (c)
Explanation: Phenol is a weak acid, but it is weaker than carbonic acid ($H_{2}CO_{3}$). Phenols do not react with sodium bicarbonate to release $CO_{2}$, a test used to distinguish them from stronger acids like $H_{2}CO_{3}$ or $RCOOH$.
31. An ether is more volatile than an alcohol having the same molecular formula due to
- (a) Dipolar character of ethers
- (b) Alcohols having resonance
- (c) Intermolecular H-bonding in ethers
- (d) Intermolecular H-bonding in alcohols
Correct Answer: (d)
Explanation: Alcohols possess intermolecular hydrogen bonding, which significantly increases their boiling points. Ethers lack this strong attraction between molecules, making them more volatile (lower boiling points) than isomeric alcohols.
32. Butanoic acid $\xrightarrow{PCl_{5}}$ [A] $\xrightarrow{H_{2}, Pd/BaSO_{4}, S}$ [B]. Product B is
- (a) [structure a]
- (b) [structure b] butanal
- (c) [structure c]
- (d) [structure d]
Correct Answer: (b)
Explanation: Butanoic acid reacts with $PCl_{5}$ to form butanoyl chloride. Reaction with $H_{2}/Pd-BaSO_{4}$ (Rosenmund reduction) converts the acid chloride into butanal (butyraldehyde).
33. Which of the following will not give iodoform test?
- (a) Acetophenone
- (b) Acetone
- (c) Ethanol
- (d) Acetyl chloride
Correct Answer: (d)
Explanation: The iodoform test is given by compounds with a methyl ketone group ($CH_{3}CO-$) or a methyl carbinol group ($CH_{3}CH(OH)-$). While acetophenone, acetone, and ethanol meet these criteria, acetyl chloride ($CH_{3}COCl$) does not respond to this test.
34. Which of the following is a cationic detergent?
- (a) Sodium lauryl sulphate
- (b) Sodium stearate
- (c) Cetyltrimethyl ammonium bromide
- (d) Sodium dodecylbenzene sulphonate
Correct Answer: (c)
Explanation: Cetyltrimethyl ammonium bromide is a quaternary ammonium salt, making it a cationic detergent. The others are anionic detergents.
35. Anisole $\xrightarrow{(CH_{3})_{3}CCl, AlCl_{3}}$ [X]. Product X is
- (a) p-tert-butylanisole
- (b) [structure b]
- (c) [structure c]
- (d) [structure d]
Correct Answer: (a)
Explanation: Anisole undergoes Friedel-Crafts alkylation. The methoxy group ($-OCH_{3}$) is strongly activating and ortho/para directing. Due to the bulkiness of the tert-butyl group, the para-isomer (p-tert-butylanisole) is the major product.
36. Pollutant of automobile exhausts that affects nervous system and mental health is
- (a) lead
- (b) mercury
- (c) nitric oxide
- (d) sulphur dioxide
Correct Answer: (b)
Explanation: While lead is a common exhaust pollutant, Mercury is noted for its severe impacts on the nervous system and cognitive health. (Note: Lead is also highly neurotoxic, but according to the provided key, choice (b) is selected).
37. Organic compound A + $NH_{3} \rightarrow$ B $\xrightarrow{\Delta}$ C $\xrightarrow{Br_{2}/KOH}$ $CH_{3}CH_{2}NH_{2}$. A is
- (a) $CH_{3}CH_{2}COOH$
- (b) $CH_{3}COOH$
- (c) $CH_{3}CH_{2}CH_{2}COOH$
- (d) isobutyric acid
Correct Answer: (a)
Explanation: The final product is ethylamine ($C_{2}$). The Hofmann bromamide reaction (step $C \rightarrow ethylamine$) reduces the chain by one carbon. Thus, C must be propanamide ($C_{3}$). C is formed by heating the ammonium salt B of acid A. Therefore, A must be propanoic acid ($C_{3}$, $CH_{3}CH_{2}COOH$).
38. Synthetic human hair wigs are made from a copolymer of vinyl chloride and acrylonitrile called
- (a) PVC
- (b) polyacrylonitrile
- (c) cellulose
- (d) dynel
Correct Answer: (d)
Explanation: Dynel is the trade name for a synthetic copolymer of vinyl chloride and acrylonitrile often used in wigs [This is a common polymer application fact].
39. Glycogen is a branched polymer of
- (a) $\alpha$-D-glucose
- (b) $\beta$-D-glucose
- (c) $\alpha$-D-fructose
- (d) none
Correct Answer: (a)
Explanation: Glycogen, known as animal starch, is a highly branched polymer composed of $\alpha$-D-glucose units.
40. Which statement about sucrose is incorrect?
- (a) It contains glucose in furanose form and fructose in pyranose.
- (b) It forms an octaacetate.
- (c) It is a non-reducing sugar.
- (d) Hydrolysis gives invert sugar.
Correct Answer: (a)
Explanation: In sucrose, glucose exists in pyranose form (6-membered) and fructose exists in furanose form (5-membered). Statement (a) describes the opposite.
41. For the reaction, $CH_{3}CHO + HCN \rightarrow$ [Product]. The product is a
- (a) racemic mixture of 1:1 enantiomers
- (b) mixture of 1:1 diastereomers
- (c) mixture of 1:2 enantiomers
- (d) mixture of 1:1 enantiomers of aldehyde
Correct Answer: (a)
Explanation: The addition of $HCN$ to acetaldehyde creates a chiral center. Since the starting material is achiral and the environment is symmetrical, equal amounts of both enantiomers are formed, resulting in a racemic mixture [Standard carbonyl addition logic].
42. Which suggested test distinguishes the given pair?
- (a) $1^{\circ}$ and $2^{\circ}$ amine (carbylamine test)
- (b) $CH_{3}CHO$ and $CH_{3}CH_{2}CHO$ (Tollen's test)
- (c) $CH_{3}OH$ and $CH_{3}CH_{2}OH$ (Lucas test)
- (d) $CH_{3}COCH_{3}$ and $CH_{3}CH_{2}COCH_{3}$ (Brady's reagent)
Correct Answer: (a)
Explanation: The carbylamine test is specific to primary ($1^{\circ}$) amines. They produce a foul-smelling isocyanide, while secondary ($2^{\circ}$) and tertiary amines do not react.
43. Identify (X) in: $C_{3}H_{8}O \xrightarrow{K_{2}Cr_{2}O_{7}/H_{2}SO_{4}}$ $C_{3}H_{6}O \xrightarrow{I_{2}/NaOH}$ $CHI_{3}$
- (a) n-propanol
- (b) Isopropanol
- (c) Ethyl methyl ether
- (d) Propanal
Correct Answer: (b)
Explanation: Since the oxidation product ($C_{3}H_{6}O$) gives a positive iodoform test ($CHI_{3}$), it must be acetone ($CH_{3}COCH_{3}$). Acetone is obtained by the oxidation of isopropanol (2-propanol). n-Propanol would oxidize to propanal, which does not give the iodoform test.
44. Among the following, elastomers are: (1) natural rubber, (2) bakelite, (3) Buna-S, (4) dacron.
- (a) 2 and 4
- (b) 3 and 2
- (c) 1 and 4
- (d) 1 and 3
Correct Answer: (d)
Explanation: Natural rubber and Buna-S (SBR) are both elastomers. Bakelite is a thermosetting polymer and Dacron is a fibre.
45. Among the following, the strongest acid is
- (a) $CH_{3}COOH$
- (b) $CH_{2}ClCH_{2}COOH$
- (c) $CH_{2}ClCOOH$
- (d) $CH_{3}CH_{2}COOH$
Correct Answer: (c)
Explanation: Chloroacetic acid ($CH_{2}ClCOOH$) is stronger than acetic acid due to the electron-withdrawing inductive ($-I$) effect of Chlorine. It is also stronger than $\beta$-chloropropanoic acid (b) because the inductive effect decreases rapidly with distance from the carboxyl group.