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

Hint: 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$$3$$.