wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

2.2 g of an alcohol (A) when treated with CH3-MgI liberates 560 mL of CH4 at STP. Alcohol (A) on dehydration followed by ozonolysis gives ketone (B) along with (C). Oxime of ketone (B) contains 19.17% N. (A) on oxidation gives ketone (D) having the same number of the carbon atom.
The molecular mass of (A) is?

A
74
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
88
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
60
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
102
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is B 88
ROH+CH3MgICH4+ROMgI
Number of moles of CH4 = number of moles of alcohol
56022400=2.2M=M=88 [where M = molecular weight]
Thus molecular formula of alcohol A is C5H11OH.

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Chemical Properties of Alcohols 1
CHEMISTRY
Watch in App
Join BYJU'S Learning Program
CrossIcon