If optical rotation produced by the compound(A) is +52 $^{o}$ , the one produced by compound (B) is :

- 52

^{o}

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

+ 52

^{o}

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

0

^{o}

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

unpredictable

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

Open in App

Solution

The correct option is C $- 52$ $^{o}$
Since (A) and (B) are enantiomers, so specific rotation of B is $- 52^{o}$ (because enantiomers have equal and opposite specific rotation)

Suggest Corrections

Similar questions

If optical rotation produced by the compound (A) is $+ 52^{\circ}$ , the one produced by compound (B) is:

cos 38^{o} cos 52^{o} - sin 38^{o} sin 52^{o} = 0

cos 38^{o} cos 52^{o} - sin 38^{o} sin 52^{o} = 0

+ 65^{0}

cos 38^{o} \times cos 52^{o} - sin 38^{o} \times sin 52^{o}

Join BYJU'S Learning Program