Question

How many ketones are possible with the molecular formula${C_6}{H_{12}}O$ ?
A) $5$
B) $6$
C) $7$
D) $8$

Answer 1

To answer the question we need to know how we can draw structures from a given molecular formula. Utilizing evenness, and the way that it should be a ketone, the carbonyl group can just move between carbons $2$ and$3$. That makes for the initial two isomers. Two additional isomers include moving the terminal methyl bunch up one to carbon$4$. Two additional isomers include moving the terminal methyl bunch up one to carbon$4$. The fifth and 6th isomers are enantiomers, the highest point of which is $R$ and the lower part of which is$S$. The thought was to move the methyl bunch up one more to carbon$3$, making a sec-butyl bunch off of${(C{H_3})_2}CO$. The seventh isomer was essentially to make a tert-butyl bunch hanging off of${(C{H_3})_2}CO$.

Complete step-by-step answer:
The simplest complex ketone is$C{H_3} - C( = O) - C{H_3}$. Its atomic formula is${C_3}{H_6}O$. From this formula, we can say that for "$n$" carbon iotas we need "$2n$" hydrogen molecules and an oxygen particle. Henceforth the overall recipe of the ketone is${C_n}{H_{2n}}O$.

Keto group contains a carboxyl gathering which has two alkyl bunches connected to it each on one or the other side. Primary formula: $R - {C = O} - {R^1}$. Aldehydes and ketones are natural mixes that join a carbonyl functional group,$C = O$.

We can write seven ketone structures with the molecular formula${C_6}{H_{12}}O$.



7)

Note: We know that a carbonyl group can only exist at spots that give a $s{p^2}$ hybridization, so there is no structure for the carbonyl on carbon $3$ when carbon $3$ has a tert-butyl group. Similarly, there is no structure for which the carbonyl is on carbon $3$ while the sec-butyl is also on carbon$3$.