Compounds P, Q, R and S below all have the same molecular formula, C4H8. - P: CH2=CHCH2CH3 - Q: CH3CH=CHCH3 - R: (CH3)2C=CH2 - S: cyclobutane, a four-membered ring of carbon atoms joined only by single C-C bonds (no C=C double bond) (a) State the general formula, in terms of n,…

Answer: ((a)) general formula CnH2n; for n=4, 2n=8, giving C4H8, consistent with P, Q and R ((b)) 2-methylprop-1-ene (compound R); the C=C double bond (the alkene functional group) is present in P, Q and R but completely absent from S, which contains only single C-C bonds ((c)(i)) positional isomerism; P and Q share the same unbranched four-carbon skeleton and molecular formula C4H8, but the C=C double bond occupies a different position (C1-C2 in P, C2-C3 in Q) ((c)(ii)) chain isomerism; P and R share molecular formula C4H8 and both have a terminal C=C double bond, but the carbon skeleton itself differs. P is an unbranched four-carbon chain, R is a branched chain (a three-carbon chain with a methyl branch) ((c)(iii)) functional group isomerism; P and S share molecular formula C4H8, but P contains a C=C double bond (the alkene functional group) while S (cyclobutane) has no C=C double bond at all, only single bonds arranged in a ring ((d)) each carbon of the C=C double bond must carry two different groups; only Q satisfies this at both double-bond carbons, so only Q shows cis-trans isomerism. P fails at C1 (two identical H atoms), and R fails at both double-bond carbons (two identical H atoms on one, two identical…. Part (a): General formula of the alkenes Alkenes containing one C=C double bond and no ring form the homologous series with general formula: CnH2n For n=4, 2n=8, giving C4H8. Exactly the molecular formula given for P, Q and R, confirming a…

AS Chemistry 9701: Naming Alkenes and Chain, Positional, Functional Group Isomerism

Compounds P, Q, R and S below all have the same molecular formula, $\text{C}_4\text{H}_8$ .

P: $\text{CH}_2=\text{CHCH}_2\text{CH}_3$
Q: $\text{CH}_3\text{CH}=\text{CHCH}_3$
R: $(\text{CH}_3)_2\text{C}=\text{CH}_2$
S: cyclobutane, a four-membered ring of carbon atoms joined only by single C-C bonds (no C=C double bond)

(a) State the general formula, in terms of $n$ , of the homologous series of non-cyclic alkenes containing one C=C double bond, and use it to confirm that $\text{C}_4\text{H}_8$ is consistent with compounds P, Q and R. [2]

(b) Give the IUPAC name of compound R, and state the type of functional group present in P, Q and R that is completely absent from compound S. [2]

(c) State the type of structural isomerism (chain, positional or functional group) shown by each of the following pairs of compounds, giving a brief reason for each answer:

(i) P and Q [1]

(ii) P and R [1]

(iii) P and S [1]

(d) State the structural condition that must be satisfied at each carbon atom of a C=C double bond for cis-trans (geometrical) isomerism to be possible, and use it to determine, with a reason, which one of P, Q and R can show this type of isomerism. [2]

Part (a): General formula of the alkenes

Alkenes containing one C=C double bond and no ring form the homologous series with general formula:

$\text{C}_n\text{H}_{2n}$

For $n=4$ , $2n=8$ , giving $\text{C}_4\text{H}_8$ . Exactly the molecular formula given for P, Q and R, confirming all three are consistent with a single-double-bond, non-cyclic alkene of four carbons.

Part (b): Naming R and its functional group

$(\text{CH}_3)_2\text{C}=\text{CH}_2$

The longest carbon chain that includes both carbons of the C=C double bond runs through only three carbons (propene), not four: writing it as $\text{CH}_2=\text{C(CH}_3\text{)CH}_3$ , the third carbon of this propene chain is itself a methyl group, and there is a second methyl group branching off C2. Numbering from the $=\text{CH}_2$ end gives the double bond the lowest possible locant (C1=C2), with the methyl branch on C2. This gives the IUPAC name:

$\textbf{2-methylprop-1-ene}$

P, Q and R all contain a C=C double bond, the alkene functional group. Compound S (cyclobutane) contains no C=C double bond at all; its four carbons are joined only by single bonds in a ring, so this functional group is completely absent from S.

Part (c): Classifying the isomerism between each pair

(i) P and Q

$\text{CH}_2=\text{CHCH}_2\text{CH}_3 \ \text{(P)} \qquad \text{CH}_3\text{CH}=\text{CHCH}_3 \ \text{(Q)}$

Both molecules have molecular formula $\text{C}_4\text{H}_8$ and the same unbranched, four-carbon skeleton. What differs is where the C=C double bond sits along that chain (between C1–C2 in P, C2–C3 in Q). Because the carbon skeleton is identical and only the position of the double bond changes, this is positional isomerism.

(ii) P and R

$\text{CH}_2=\text{CHCH}_2\text{CH}_3 \ \text{(P)} \qquad (\text{CH}_3)_2\text{C}=\text{CH}_2 \ \text{(R)}$

Both molecules have molecular formula $\text{C}_4\text{H}_8$ and both have a C=C double bond at the very start of a chain (a terminal alkene). Here, however, the carbon skeleton itself is different: P is an unbranched chain of four carbons, while R is a branched skeleton, a three-carbon chain with a methyl group branching off the double-bond carbon. Because the underlying skeleton differs (straight vs branched), rather than just the position of the functional group on an identical skeleton, this is chain isomerism.

(iii) P and S

$\text{CH}_2=\text{CHCH}_2\text{CH}_3 \ \text{(P)} \qquad \text{cyclobutane (S)}$

P and S both have molecular formula $\text{C}_4\text{H}_8$ , but they contain different functional groups: P has a C=C double bond, while S has no double bond anywhere (its four carbons are simply joined in a ring by single bonds. Because the two compounds share a molecular formula but differ in the functional group present (double bond vs ring, no double bond), this is functional group isomerism) the standard relationship between an alkene and a cycloalkane of the same molecular formula.

Part (d): Condition for cis-trans isomerism, and which compound shows it

Cis-trans (geometrical) isomerism arises because rotation about a C=C double bond is restricted. However, this restricted rotation only produces genuinely different spatial arrangements if each carbon of the double bond is attached to two different groups. If either double-bond carbon carries two identical groups, swapping their positions makes no difference, and only one spatial arrangement exists.

In P ( $\text{CH}_2=\text{CHCH}_2\text{CH}_3$ ), the first double-bond carbon (C1) is a $=\text{CH}_2$ group, it carries two identical hydrogen atoms. C1 fails the condition, so P cannot show cis-trans isomerism, regardless of what is attached to C2.
In Q ( $\text{CH}_3\text{CH}=\text{CHCH}_3$ ), C2 carries a methyl group and a hydrogen atom (different), and C3 carries a methyl group and a hydrogen atom (different). Both double-bond carbons satisfy the condition, so Q can exist as two distinguishable stereoisomers (the familiar cis- and trans-but-2-ene).
In R ( $(\text{CH}_3)_2\text{C}=\text{CH}_2$ ), the $=\text{CH}_2$ carbon carries two identical hydrogen atoms, and the other double-bond carbon carries two identical methyl groups. R fails the condition at both carbons, so it cannot show cis-trans isomerism either.

Of the three, only Q shows cis-trans isomerism.

Final answers

(a) $\text{C}_n\text{H}_{2n}$ ; consistent with $\text{C}_4\text{H}_8$ since $2(4)=8$
(b) 2-methylprop-1-ene; the C=C double bond is present in P, Q, R but absent from S
(c) (i) Positional isomerism (P, Q); (ii) Chain isomerism (P, R); (iii) Functional group isomerism (P, S)
(d) Each double-bond carbon needs two different groups; only Q (but-2-ene) shows cis-trans isomerism. P and R each fail the condition on at least one carbon