Optical Activity & Numericals

CHEStereochemistry & Optical Activity

A molecule with a chiral (asymmetric) carbon is optically active — it rotates plane-polarised light and exists as two non-superimposable mirror images. Numericals here use specific rotation and the make-up of enantiomeric mixtures.

Specific rotation

[\alpha] = \dfrac{\alpha}{l \times c}

\alpha

= observed rotation in degrees,

l

= path length in dm,

c

= concentration in

g/mL

Number of optical isomers

N = 2^n

n

= number of dissimilar chiral carbons; a meso form reduces the active count

Worked example (sample numerical)

[\alpha] = \dfrac{+5.0^\circ}{1\,\text{dm} \times 0.20\,g/mL} = +25^\circ\,mL\,g^{-1}\,dm^{-1}

c = 2\,g/10\,mL = 0.20\,g/mL

A chiral carbon is an $sp^3$ carbon bonded to four different groups; the molecule then has no plane of symmetry and is optically active.
Enantiomers rotate light equally but in opposite senses — (+)/dextro and (−)/laevo — and are identical in all other physical properties.
A racemic mixture ( $\pm$ ) is a 50:50 pair of enantiomers, so its net rotation is zero (external compensation).
$S_N1$ at a chiral centre gives racemisation; $S_N2$ gives inversion (a single, inverted enantiomer).
Specific rotation: $[\alpha] = \dfrac{\alpha}{l \times c}$ , where $\alpha$ is the observed rotation, $l$ the path length in dm and $c$ the concentration in g/mL.
Number of optical isomers $= 2^n$ for $n$ dissimilar chiral centres; a meso compound has an internal plane of symmetry and is optically inactive.
Enantiomers differ only in the direction they rotate plane-polarised light and in their reaction with other chiral substances; melting point, boiling point and solubility are identical.
Among $C_4H_9Br$ isomers, only 2-bromobutane is chiral (its $C2$ bears H, Br, $CH_3$ and $C_2H_5$ ); 1-chlorobutane, 2-chloropropane and similar lack a carbon with four different groups.
The sign of rotation, $(+)$ or $(-)$ , is determined experimentally and is not predictable from the D/L or R/S configuration label.
Racemisation in $S_N1$ arises because the planar carbocation is attacked from both faces, giving roughly equal amounts of the two enantiomers (often a slight excess of inversion).
In a polarimeter, $l$ is taken in decimetres (dm) and $c$ in $g/mL$ ; using cm or $g/L$ without conversion gives a wrong specific rotation.
A meso compound contains chiral centres but is achiral overall due to an internal mirror plane, so two-stereocentre systems often show three stereoisomers (a $d$ , an $l$ and a meso) rather than four.

Deciding whether a molecule is optically active

1Find every $sp^3$ carbon and check whether any one carries four different groups. That carbon is the chiral (asymmetric) centre.
2Look for a plane of symmetry in the whole molecule. If one exists the molecule is achiral even when it contains chiral centres — that is the meso case.
3Count the chiral centres, $n$ . The maximum number of optical isomers is $2^{n}$ , reduced below that whenever a meso form exists.
4A racemic mixture contains the two enantiomers in equal amounts, so the rotations cancel exactly and the mixture is optically inactive — inactive by external compensation, which is the phrase the marking scheme wants.

Where the marks go

Using the path length in cm instead of dm, or concentration in $g/L$ instead of $g/mL$ , in the specific-rotation formula.
Calling a racemic mixture optically active; its two enantiomers cancel, giving zero net rotation.
Forgetting that a meso compound, despite having chiral carbons, is optically inactive (internal compensation).
Predicting the sign of rotation from R/S or D/L labels — the $(+)/(-)$ sign must be measured, not deduced.
Assigning $S_N1$ as giving inversion or $S_N2$ as giving racemisation; $S_N1$ racemises and $S_N2$ inverts at the chiral centre.

How the board asks it

Numericalspecific-rotation formula
A $2\,\text{dm}$ polarimeter tube is filled with a solution containing $0.5\,\text{g}$ of a sugar in $10\,\text{mL}$ of water. If the observed rotation is $+6.6^\circ$ , calculate the specific rotation of the sugar.
Numericalunit conversion of $l$ and $c$
A student recorded the path length as $20\,\text{cm}$ and the concentration as $50\,\text{g/L}$ . Convert these to the units required by the polarimeter and hence calculate $[\alpha]$ if the observed rotation is $+2.5^\circ$ .
Numericalnumber of optical isomers $= 2^n$
A compound has $3$ dissimilar chiral carbon atoms. Calculate the maximum number of optical isomers possible, and state how this number is reduced if the molecule possesses an internal plane of symmetry (a meso form).
Give reasonsracemic mixture and external compensation
Account for the fact that a racemic mixture of $2$ -bromobutane shows no rotation of plane-polarised light, even though each of its components is optically active.
Assertion–Reason $S_N1$ racemisation via a planar carbocation
Assertion: Hydrolysis of optically active $2$ -bromobutane by an $S_N1$ pathway gives an optically inactive product. Reason: The planar carbocation intermediate is attacked with equal probability from both faces. State whether each statement is true and whether the reason correctly explains the assertion.

Practise this topic

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Written for Sublevo. Question text quoted anywhere in these notes is the Council’s and carries its year and paper; the board’s own diagrams are not reproduced.