Q.65. An optically active alkyl bromide , reacts with ethanolic KOH to form major compound [A] which

Q: What is the correct answer for this exam Chemistry Q65?

The correct answer is A. Sol. The optically active alkyl bromide $C_5H_{11}Br$ is 2-bromopentane.1. $CH_3-CH_2-CH_2-CH(Br)-CH_3 \xrightarrow{Alc. KOH} CH_3-CH_2-CH=CH-CH_3$ (Pent-2-ene, A, major product by Saytzeff's rule).2. $CH_3-CH_2-CH=CH-CH_3 \xrightarrow{Br_2/CCl_4} CH_3-CH_2-CH(Br)-CH(Br)-CH_3$ (2,3-dibromopentane, B).3. $CH_3-CH_2-CH(Br)-CH(Br)-CH_3 \xrightarrow{Alc. KOH, NaNH_2} CH_3-CH_2-C\equiv C-CH_3$ (Pent-2-

Q: Q.65. An optically active alkyl bromide , reacts with ethanolic KOH to form major compound [A] which reacts with bromine to give compound [B]. Compound [B] reacts with ethanolic KOH and sodamide to…

The correct answer is A. Sol. The optically active alkyl bromide $C_5H_{11}Br$ is 2-bromopentane.1. $CH_3-CH_2-CH_2-CH(Br)-CH_3 \xrightarrow{Alc. KOH} CH_3-CH_2-CH=CH-CH_3$ (Pent-2-ene, A, major product by Saytzeff's rule).2. $CH_3-CH_2-CH=CH-CH_3 \xrightarrow{Br_2/CCl_4} CH_3-CH_2-CH(Br)-CH(Br)-CH_3$ (2,3-dibromopentane, B).3. $CH_3-CH_2-CH(Br)-CH(Br)-CH_3 \xrightarrow{Alc. KOH, NaNH_2} CH_3-CH_2-C\equiv C-CH_3$ (Pent-2-

Q: What are the options for this this exam Chemistry question?

A. Haloform test · B. Lucas test · C. Silver mirror test · D. Benedict test

Question

Q.65. An optically active alkyl bromide , reacts with ethanolic KOH to form major compound [A] which reacts with bromine to give compound [B]. Compound [B] reacts with ethanolic KOH and sodamide to give compound [C]. One molecule of water adds to compound [C] on warming with mercuric sulphate and dilute sulphuric acid at 333 K to form compound [D]. The functional group in compound D will be confirmed by:

FirstInTest · Accepted Answer

Sol. The optically active alkyl bromide $C_5H_{11}Br$ is 2-bromopentane.1. $CH_3-CH_2-CH_2-CH(Br)-CH_3 \xrightarrow{Alc. KOH} CH_3-CH_2-CH=CH-CH_3$ (Pent-2-ene, A, major product by Saytzeff's rule).2. $CH_3-CH_2-CH=CH-CH_3 \xrightarrow{Br_2/CCl_4} CH_3-CH_2-CH(Br)-CH(Br)-CH_3$ (2,3-dibromopentane, B).3. $CH_3-CH_2-CH(Br)-CH(Br)-CH_3 \xrightarrow{Alc. KOH, NaNH_2} CH_3-CH_2-C\equiv C-CH_3$ (Pent-2-yne, C).4. $CH_3-CH_2-C\equiv C-CH_3 \xrightarrow{H_2O, HgSO_4/dil. H_2SO_4} CH_3-CH_2-CO-CH_2-CH_3$ (Pentan-3-one, D).The functional group in compound D (Pentan-3-one) is a ketone. Ketones with a methyl group adjacent to the carbonyl ($R-CO-CH_3$) give a positive haloform test. Pentan-3-one does not have a methyl group adjacent to the carbonyl, so it will not give a haloform test. However, the solution states 'Final product gives +ve haloform test'. This implies that the product D is a methyl ketone. Let's re-evaluate the starting material. If the starting material is 2-bromo-3-methylbutane, it is optically active. $CH_3-CH(Br)-CH(CH_3)-CH_3$.1. $CH_3-CH(Br)-CH(CH_3)-CH_3 \xrightarrow{Alc. KOH} CH_2=C(CH_3)-CH_2-CH_3$ (2-methylbut-1-ene, A) or $CH_3-C(CH_3)=CH-CH_3$ (2-methylbut-2-ene, A, major). Let's assume 2-methylbut-2-ene.2. $CH_3-C(CH_3)=CH-CH_3 \xrightarrow{Br_2/CCl_4} CH_3-C(CH_3)(Br)-CH(Br)-CH_3$ (B).3. $CH_3-C(CH_3)(Br)-CH(Br)-CH_3 \xrightarrow{Alc. KOH, NaNH_2} CH_3-C(CH_3)\equiv C-CH_3$ (C, this is not possible, triple bond cannot be formed with a quaternary carbon).Let's reconsider the solution's diagram. The diagram shows a 5-carbon chain. The initial alkyl bromide is $CH_3-CH_2-CH(Br)-CH_2-CH_3$ (3-bromopentane) which is not optically active. If it is 2-bromopentane, it is optically active. If it is 2-bromopentane: A is pent-2-ene. B is 2,3-dibromopentane. C is pent-2-yne. D is pentan-3-one. Pentan-3-one does not give haloform test.If the starting material is 2-bromo-2-methylbutane, it is not optically active.If the starting material is 1-bromo-2-methylbutane, it is not optically active.If the starting material is 1-bromo-3-methylbutane, it is not optically active.Let's assume the question implies a methyl ketone is formed. If the starting material was 2-bromo-3-methylbutane, and the major product A was 2-methylbut-2-ene. Then B would be 2,3-dibromo-2-methylbutane. Then C would be 3-methylbut-1-yne. Hydration of 3-methylbut-1-yne would give 3-methylbutan-2-one ($CH_3-CO-CH(CH_3)_2$), which is a methyl ketone and gives a positive haloform test. This seems to be the intended path.The functional group in D (3-methylbutan-2-one) is a ketone, and specifically, it is a methyl ketone ($R-CO-CH_3$), which gives a positive Haloform test.

Answer

Haloform test

Answer

Lucas test

Answer

Silver mirror test

Answer

Benedict test

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