Isomerism: structural isomerism and stereoisomerism — practice questions

The molecular formula C₄H₁₀ corresponds to an alkane ($C_nH_{2n+2}$, where n=4).

Step 1: Draw the straight-chain isomer. First, arrange the four carbon atoms in a continuous chain and add hydrogen atoms to satisfy the valency of each carbon. CH₃-CH₂-CH₂-CH₃ This is butane.

Step 2: Draw branched-chain isomers. Try arranging the carbons in a different way. We can have a main chain of three carbons with one carbon as a branch. CH₃-CH(CH₃)-CH₃ Place the branch on the central carbon (placing it on an end carbon would just result in butane again). This is 2-methylpropane.

Step 3: Check for other possibilities. There are no other ways to arrange 4 carbon atoms. Therefore, there are two structural isomers of C₄H₁₀: butane and 2-methylpropane. These are examples of chain isomerism.

The structure is CH₃CH=CHCH₃. Let's analyse the groups on the C=C double bond.

Step 1: Identify the groups on each carbon of the C=C bond.

• The left carbon is bonded to an H atom and a CH₃ group.
• The right carbon is also bonded to an H atom and a CH₃ group. Since each carbon has two different groups, E/Z isomerism is possible.

Step 2: Assign priorities using CIP rules. On each carbon, the CH₃ group has a higher priority than the H atom because carbon (atomic number 6) has a higher atomic number than hydrogen (atomic number 1).

Step 3: Draw the Z-isomer. The Z-isomer has the high-priority groups (CH₃) on the same side of the double bond. Skeletal formula for Z-but-2-ene:

  /\ 
 /  \ 
C====C

(Imagine the two methyl groups pointing 'up' from the double bond).

Step 4: Draw the E-isomer. The E-isomer has the high-priority groups (CH₃) on opposite sides of the double bond. Skeletal formula for E-but-2-ene:

  / 
 /   \
C====C
     /
    / 

(Imagine one methyl group pointing 'up' and the other 'down' from the double bond).