In the reaction

$C H_{3} C H O + H C N \to C H_{3} - C H (O H) C N H + / H_{2} O - ------- \to C H_{3} - C H (O H) C O O H$

An asymmetric centre is generated. The acid obtained would be

d- isomer

No worries! We‘ve got your back. Try BYJU‘S free classes today!

l - isomer

No worries! We‘ve got your back. Try BYJU‘S free classes today!

50% d + 50%l - isomer

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

20% d + 80% l - isomer

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C
50% d + 50%l - isomer

A racemic mixture is always produced whenever a new chiral carban atom is generated in a Reaction. Thus 50 $%$ d and 50 $%$ l isomers are produced.

Suggest Corrections

Similar questions

C H_{3} C H O + H C N \to C H_{3} C H (O H) C N H . O H - --- \to C H_{3} C H (O H) C O O H

C H_{3} C H O + H C N \to C H_{3} C H (O H) C N H^{+} - - \to H_{2} O C H_{3} C H (O H) C O O H

[Pt(NH₃)₄Cl₂]Br₂ and [Pt(NH₃)₄Br₂]Cl₂ is a type of:

C H_{3} - C H_{3} | C | C H_{3} -

C H_{2} C H_{2} O H K_{2} C r_{2} O_{7}; H_{2} S O_{4} \to H_{2} O, h e a t A S O C l_{2} \to B (C H_{3})_{2} N H \to 2 m o l e C 1. L i A l H_{4} d i e t h y l e t h e r \to 2. H_{2} O D

[C o C l_{2} (N H_{3})_{4}]^{+} + C l^{-} \to [C o C l_{3} (N H_{3})_{3}] + N H_{3}

Join BYJU'S Learning Program