Part 5: STRUCTURE DETERMINATION

a. For a hydrocarbon, the formula for the index of hydrogen defficiency, IHD = 0.5 ( 2c+2-h), therefore, the IHD = 0.5 (2 x 8 + 2 -14) = 2
FYI : The IHD = pi + r (pi bonds plus rings) there can be 2 units of unsaturation, which could mean 2 rings, a ring and a double bond, two double bonds or a triple bond.

b. Here are possible answers (there are likely others), these were selected to represent four different ways to get an IHD=2. Note that there has to be some symmetry.

c. The structure shown below actually has two E double bonds (the higher priority alkyl groups need to be on opposite sides of the C=C).

d. The IR data 2150 cm-1 implies an alkyne, the 3300 cm-1 implies an sp C-H, i.e. terminal alkyne. A generic example and a specific example are shown below. Note that alkyne geometry in linear.

e.
i.
The terminal alkyne will be the most acidic choice, pKa about 25.
ii
The conjugate base of an alkyne has the -ve charge associated with an sp hybrid orbital. This means the electrons are in a small orbital (due to the high s character) that puts the -ve charge closer to the positive nucleus. This stabilises the conjugate base. (i.e. think of HA <=> H+ and A-, stablising A- favours the right hand side indicating that HA is a stronger acid).
iii. A generic balanced equation for the reaction of a terminal alkyne as an acid is shown.:

f.
i. The reaction will be allylic radical bromination. This is the expected product because allylic C-H bonds are the weakest and this leads to a resonance stablised radical.

ii. Comparing the initial molecular formula with C8H14Br2 indicates that an addition reaction has occurred (addition to a C=C as per the hydrocarbon experiment from the laboratory), likely if a high concentration of bromine was used (in error).

Common errors:
Incorrectly counting the types of H.
Ignoring the fact that the question indicates that X was a hydrocarbon.
Did not recognise the terminal alkyne in part d.
Did not use a strong enough base (pKa >25) to deprotonate the terminal alkyne.
Confused the pKa of internal and terminal alkynes.
Drawing vinyl bromides (C=C-Br) as products of radical bromination (it's a reaction of sp3 C-H bonds).
Not recognising the electrophilic addition in part fii.