Isomerism

The compound has two double bonds:

$ \mathrm{CH}_3-\mathrm{CH}=\mathrm{CH}-\mathrm{CH}=\mathrm{CH}_2 $

Only the central $\mathrm{C}=\mathrm{C}$ (second one) can show geometrical isomerism, as the terminal alkene ( $\mathrm{CH}=\mathrm{CH}_2$ ) lacks two different groups on one carbon.

So, only one C=C shows cis-trans isomerism $\rightarrow 2$ isomers

There are total 5 isomers (including stereoisomers) of compound having formula $ \mathrm{C}_{4} \mathrm{H}_{9} \mathrm{Br} $

1-bromobutane $ \rightarrow $

$ \mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{Br} $

2-bromobutane $ \rightarrow $

( $ d $ and $ l $ form)

1-bromo-2-methyl propane $ \rightarrow $

2-bromo-2-methyl propane $ \rightarrow $

(Molecular weight of $186 u$ is of dibromoethene. It can have 3 isomers, two geometrical isomers (cis and trans) of 1,2 dibromoethene and one $ \text { 1-1-dibromoethene. } $

Threonine is the amino acid which is consisted of two chiral carbons.

Chiral carbons are those carbon atoms which are attached to four different substituent (asymmetric).

An optically active compounds exists in two isomeric form that rotate the plane polarised light in opposite direction. This phenomenon is called optical isomerism. Among the given complexes, $\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_3\left(\mathrm{NO}_2\right)_3\right]$ does not show optical isomerism, because it has symmetry in its structure.

Although $ \mathrm{CH}_{3} \mathrm{C}=\mathrm{CH}-\mathrm{CH}=\mathrm{CH}_{2} $ has two double bond but one of them has two similar atoms (i.e. H -atom) therefore no geometrical isomerism is possible around the double bond. However each carbon atom of other double bond has two different atoms/groups and hence geometrical isomerism is possible around this double bond.

Hence, only two geometrical isomers are possible.

If two compounds have the same empirical formula but different molecular formulae, they must have different molecular weights. The empirical formula represents the simplest whole-number ratio of the elements in the compound, but it doesn't provide information about the actual number of atoms in a molecule (molecular formula). Because the molecular formula is a multiple of the empirical formula, compounds with different molecular formulae will have different molecular weights even if their empirical formula is the same.

To illustrate with an example, consider the empirical formula $CH_2$. This empirical formula can correspond to several different molecular formulae such as $C_2H_4$, $C_3H_6$, $C_4H_8$, and so on. Each of these molecular formulae has a different molecular weight, even though they share the same empirical formula.

Therefore, the correct option is:

Option B: different molecular weights

Chiral molecule have at least one carbon atom which is attached to 4 different atoms or groups and it does not have an element of symmetry.

Alcohols are functional group isomers of ethers. e.g. $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{OH}$ and $\mathrm{CH}_3-\mathrm{O}-\mathrm{CH}_3$ both have same molecular formula but different functional group.

Formula of optical isomer in symmetrical structure = 2ⁿ, Here, n = number of chiral center.

There are two chiral center present in 2-bromo, 3- chloro butane.

Hence, n = 2

Number of optical isomer is = $2^n=2^2=4$

According to sequence rule, the priority order are

$\mathop {Cl}\limits_{(1)} $ > $\mathop {{C_2}{H_5}}\limits_{(2)} $ > $\mathop {C{H_3}}\limits_{(3)} $ > $\mathop H\limits_{(4)} $, so in R configuration.