
How do you name and draw 5 compounds that have the formula $ {C_7}{H_7}Br $ ?
Answer
532.8k+ views
Hint: To know the name of compounds which have the formula $ {C_7}{H_7}Br $ , we should first write the general formula of the given formula, then we will discuss the five compounds which have the given formula.
Complete answer:
Well, you know that an alkane would have the general formula $ {C_n}{H_{2n + 2}} $ , so this compound cannot be a straight-chained alkane.
That means you should try drawing alkenes and alkynes, and definitely try rings:-
$ * $ I tried a five-membered ring with two side-chain carbons, and added conjugated double bonds to decrease the number of hydrogens. That's one isomer.
$ * $ I tried drawing a six-membered ring with one side-chain carbon. When I placed a bromine on that carbon, I achieved $ {C_7}{H_7}Br $ . I moved the bromine onto the ring for another isomer, and moved that bromine around the ring for two more isomers for a total of five.
Seems like there are at least 14 though. If you want, three of them are bromene-ynes, which can be a challenge to name.
From left to right, top to bottom:
$ * $ $ 2 - methyl - bromobenzene,{\text{ }}or\;o - bromotoluene,{\text{ }}or{\text{ }}1 - bromo - 2 - methylbenzene $
$ * $ $ 3 - methyl - bromobenzene,{\text{ }}or\;m - bromotoluene,{\text{ }}or{\text{ }}1 - bromo - 3 - methylbenzene $
$ * $ $ 4 - methyl - bromobenzene,{\text{ }}or\;p - bromotoluene,{\text{ }}or{\text{ }}1 - bromo - 4 - methylbenzene $
$ * $ I'm not sure how to name this one, so I looked it up and got:
$ \begin{array}{*{20}{l}}
{5 - \left( {1 - bromoethylidene} \right) - 1,3 - cyclopentadiene} \\
\;
\end{array} $
This uses cyclopentadiene as the parent compound, and an ethylidene substituent ( $ C = \mathop C\limits^* - C{H_3} $ , with bromine on the starred carbon), and the upper-left carbon on cyclopentadiene is carbon-1; we'd move counter-clockwise.
Note:
Benzyl bromide is an organic compound with the formula $ {C_6}{H_5}C{H_2}Br $ . The molecule consists of a benzene ring substituted with a bromomethyl group. It is a colorless liquid with lachrymatory properties.
Complete answer:
Well, you know that an alkane would have the general formula $ {C_n}{H_{2n + 2}} $ , so this compound cannot be a straight-chained alkane.
That means you should try drawing alkenes and alkynes, and definitely try rings:-
$ * $ I tried a five-membered ring with two side-chain carbons, and added conjugated double bonds to decrease the number of hydrogens. That's one isomer.
$ * $ I tried drawing a six-membered ring with one side-chain carbon. When I placed a bromine on that carbon, I achieved $ {C_7}{H_7}Br $ . I moved the bromine onto the ring for another isomer, and moved that bromine around the ring for two more isomers for a total of five.
Seems like there are at least 14 though. If you want, three of them are bromene-ynes, which can be a challenge to name.
From left to right, top to bottom:
$ * $ $ 2 - methyl - bromobenzene,{\text{ }}or\;o - bromotoluene,{\text{ }}or{\text{ }}1 - bromo - 2 - methylbenzene $
$ * $ $ 3 - methyl - bromobenzene,{\text{ }}or\;m - bromotoluene,{\text{ }}or{\text{ }}1 - bromo - 3 - methylbenzene $
$ * $ $ 4 - methyl - bromobenzene,{\text{ }}or\;p - bromotoluene,{\text{ }}or{\text{ }}1 - bromo - 4 - methylbenzene $
$ * $ I'm not sure how to name this one, so I looked it up and got:
$ \begin{array}{*{20}{l}}
{5 - \left( {1 - bromoethylidene} \right) - 1,3 - cyclopentadiene} \\
\;
\end{array} $
This uses cyclopentadiene as the parent compound, and an ethylidene substituent ( $ C = \mathop C\limits^* - C{H_3} $ , with bromine on the starred carbon), and the upper-left carbon on cyclopentadiene is carbon-1; we'd move counter-clockwise.
Note:
Benzyl bromide is an organic compound with the formula $ {C_6}{H_5}C{H_2}Br $ . The molecule consists of a benzene ring substituted with a bromomethyl group. It is a colorless liquid with lachrymatory properties.
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