The major product for the following reaction is: $\mathrm{CH_2{=}CH{-}CN}$ (acrylonitrile) $\;+\;$ $\mathrm{HO{-}CH_2CH_2{-}SH}$ (2-mercaptoethanol) $\;\longrightarrow\;$ ?

(A) $\mathrm{HO{-}CH_2CH_2{-}S{-}CH_2CH_2{-}CN}$

(B) $\mathrm{HS{-}CH_2CH_2{-}O{-}CH_2CH_2{-}CN}$

(C) $\mathrm{CH_2{=}CH{-}C({=}NH){-}S{-}CH_2CH_2{-}OH}$

(D) $\mathrm{CH_2{=}CH{-}C({=}NH){-}O{-}CH_2CH_2{-}SH}$

JEE Main · 2023 · 13 Apr 2023 · Shift 2 · Chapter_pending

The major product for the following reaction is:

C H_{2} = CH - CN

(acrylonitrile)

+

HO - C H_{2} C H_{2} - SH

(2-mercaptoethanol)

⟶

(A) $HO - C H_{2} C H_{2} - S - C H_{2} C H_{2} - CN$
✓ Correct
(B) $HS - C H_{2} C H_{2} - O - C H_{2} C H_{2} - CN$
(C) $C H_{2} = CH - C (= NH) - S - C H_{2} C H_{2} - OH$
(D) $C H_{2} = CH - C (= NH) - O - C H_{2} C H_{2} - SH$

Answer: (A)

Solution

The two reagents are acrylonitrile, \ce{CH2=CH-CN}, and 2-mercaptoethanol, \ce{HO-CH2CH2-SH}. The electron-poor double bond of acrylonitrile is a Michael acceptor, and the thiol sulfur, being soft and highly polarisable, is a far better nucleophile than the hydroxyl oxygen. The sulfur therefore adds in a conjugate (thia-Michael) fashion to the terminal carbon of the alkene while the hydroxyl group is left untouched, joining the two fragments through a sulfide bridge. The major product is \ce{HOCH2CH2-S-CH2CH2-CN}, 3-(2-hydroxyethylsulfanyl)propanenitrile.

JEE Main · 2023 · 13 Apr 2023 · Shift 2 · Single correct option

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JEE Main · 2023 · 13 Apr 2023 · Shift 2 · Chapter_pending

Solution

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