An alkane F, C6H14 was produced. When reacted with bromine under UV light, F produced only 2 isomeric monobromo compounds G and H, with the formula C6H13Br. Compound H was chiral. Suggest the structures of F, G and H
My ans
To satisfy the condition that only 2 isomeric monobromo compounds are produced
alkane F should be 2,3 - dimethylbutane
however the mono bromo compounds of this alkane are not chiral. The products are 1-bromo-2,3-dimethylbutane and 2-bromo-2,3-dimethylbutane
Ans given in the book
Alkane F is 2,2,- dimethylbutane
however 3 isomeric monobromo compounds are possible
1-bromo-2,2-dimethylbutane
1-bromo-3,3-dimethylbutane
2-bromo-3,3-dimethylbutane
Is there some mistake in my reasoning?
Thanks in advance and Happy New Year :)
Originally posted by quailmaster:An alkane F, C6H14 was produced. When reacted with bromine under UV light, F produced only 2 isomeric monobromo compounds G and H, with the formula C6H13Br. Compound H was chiral. Suggest the structures of F, G and H
My ans
To satisfy the condition that only 2 isomeric monobromo compounds are produced
alkane F should be 2,3 - dimethylbutane
however the mono bromo compounds of this alkane are not chiral. The products are 1-bromo-2,3-dimethylbutane and 2-bromo-2,3-dimethylbutane
Ans given in the book
Alkane F is 2,2,- dimethylbutane
however 3 isomeric monobromo compounds are possible
1-bromo-2,2-dimethylbutane
1-bromo-3,3-dimethylbutane
2-bromo-3,3-dimethylbutane
Is there some mistake in my reasoning?
Thanks in advance and Happy New Year :)
Yes, there is a mistake in your reasoning. But the book's answer is also wrong.
You are correct that alkane F should be 2,3 - dimethylbutane (the book's answer is wrong), but you are wrong that "the mono bromo compounds of this alkane are not chiral".
Look very carefully.
yeah thanks i see it now.. the chiral carbon is the carbon not attached to the bromine ...
Originally posted by quailmaster:yeah thanks i see it now.. the chiral carbon is the carbon not attached to the bromine ...
Yes, tis quite a tricky question. Here's a (far) more challenging question that actually includes (you'll understand why, when you attempt it) the original Cambridge question :
When a mixture of the sodium salts of 2 monoprotic carboxylic acids are electrolyzed, a mixture of alkanes are produced.
R’COO- + RCOO- à R-R + R-R’ + R’R’
Electrolyzing a mixture of the sodium salts of 2 monoprotic carboxylic acids M and N produced 3 alkanes O, P and Q, which could be separated by fractional distillation. On titration, a solution containing 0.10g of acid N required 16.7cm3 of 0.10 mol/dm3 of NaOH(aq) to reach equivalence point. A gaseous sample of 0.10g of O occupied a volume of 54cm3 at a temperature of 380K and a pressure of 1atm. Complete combustion of a sample of P, yields 10.12g of CO2 and 6.21g of H2O. When Q was reacted with bromine under UV light, only and exactly 3 isomeric monobromo compounds R1, R2 and S were produced. S is optically inactive, while R1 and R2 are enantiomers of each other. Elucidate M, N, O, P, Q, R1, R2, S.