Organic Chem questions that can trick me! wahahaha

Yes, as the title mentions, below are the good old questions that the writer of the blog, myself, on revisiting my chemistry revision notes, can get it wrong the 2nd time.

SO, are you guys ready to take on the challenging questions coming up below?

don’t worry, detailed explanations will be provided.

Which of the following compounds has the lowest pH in aqueous solution?

1) CH3CH2CH2COCL

2) CH3CH2CHBrCOOH

3) CH3CH2CH2COOH

4) CH3CH2COCH(OH)Br

Firstly – identify your functional groups

(1) has Acyl Chloride

(2) is RCOOH and haloalkane

(3) is RCOOH

(4) a good test for your structural chem skills, it contains a ketone, an alcohol and haloalkane group

Secondly, compare the functional group’s acid strength

(2) is more acidic than (3) because the electron-withdrawing-inductive-effect of (2)’s halogen atom is going to stablilise the positive charge of the carboxylate ion of (2) more than (3).

(4) is Neutral.

remember? only phenol and carboxylic acid are acidic functional groups.

(1) is the correct answer.

RCOCL in water, undergoes hydrolysis in the aqueous solution stated in the question to form RCOOH and HCL.

HCL is without doubt, much stronger than either (2) or (3)!

The electron-donating inductive effect of an alkyl group will affect the base strength of alkylamines compared to ammonia.

Which of the following pairs correctly show the effects of the R group?

Effects on strength of RNH2                         Effect on Strength of RNH3+

as a base relative to NH3                              as an acid relative to NH4+

1)        increase                                                                          decrease

2)       decrease                                                                          decrease

3)       increase                                                                          decrease

4)      decrease                                                                           increase

First, lets get to know what a base does.

Bases are H+ acceptors

To accept H+ more readily, *notice that “readily” likes to appear in chemistry text*, a stronger base gotta have a more negative charge.

Here is where the electron-donating inductive effect of our R group come into play!

It helps to intensify the negative charge on the base, hence increasing its base strength relative to ammonia… which does not have an alkyl group.

Now to acids

Acids are H+ donors

HA ( our acid ) dissociates in water to “donate” H+

forming A-

the acids in the question may be very unfamiliar to you, so lets examine them first

RNH3+ comes from RNH2 after it has taken up a H+

RNH3+ can dissociate in water to form back RNH2 and H+

vice versa for conjugate acid NH4+ and its base partner NH3

Now, a stronger acid has to have a less negative charge on its A- (called its conjugate base) so that A- doesn’t form back HA with H+ so readily.

On the other hand,

If the negative charge on A- is intensified, by our actor again coming on stage – our alkyl group aka R, HA would forms back more readily.

There is an interesting effect to note here, for those of you following…

because compared with NH3, a stronger basic strength on RNH2…

is going to give you a weaker acid strength on its conjugate acid NH3

and this effect can be observed on all other bases and acids out there!!!

students might be confused as to why RNH2 or NH3 have basic properties even though they do not have a full negative charge.

Unlike negatively charged bases such as RCOO, which forms a covalent bond with H+ by donating its extra negatively charged electron to a very happy receiving H+…

Amines and ammonia forms a dative covalent bond with H+ instead, by sharing the 2 electron of the lone pair on N with H+.

Therefore, the resultant conjugate acid RNH3+ or NH4+ still retain the positive charge held by their H+.

As such, the answer is (1)!

3) which of the following does not have a planar structure?

A) C2O4(2-)

B) NO3(-)

C) ICL4(-)

D) PO4(3-)

let us consider C2O4(2-) first,

STEP 1 : being less electronegative than O, C is the centre atom

both Cs would be side by side in the centre, each holding the hands of 2 Os

symmetrical arrangement -> O2 – C – C – O2

STEP 2 : assign the charges

From the 2- charge, one extra electron goes to 2 lucky Oxygen atoms to give you… 2 O- atoms

And symmetrically, each carbon… gets to hold hands with 1 O- and 1 normal O atoms! 🙂

O-                                                 O-

C        –        C

O                                                O

step 3 : make the side atoms octet

O- just needs 1 more covalent bond

O needs to form 2 covalent bonds

step 4 : consider the centre atom’s electrons

Carbon is GRP IV, has 4 electrons

with 3 electrons forming 3 covalent bond with their O partners,

you are just missing 1 more electron belonging to C…

which is used to form a covalent bond between C and C

so now each carbon has 4 covalent bond, is octet

Great!

We then have to consider the shape, with 2 centre atoms ( remember step 1?) each C has 3 bond pairs and 0 lone pairs, it has a trigonal planar shape

trigonal planar = linear 🙂

Next,

NO3-

STEP 1 : N

STEP 2 : one O becomes O-

mmm, in NO3(-), notice that only 1 centre atom N

it is surrounded by 3 Os!

STEP 3 :

normal O need 2 covalent bond

normal O need 2 covalent bond

O- need 1 covalent bond

STEP 4 :

N is GRP V, has 5 electrons

BUT BUT BUT it belongs to period 2! – cannot expand octet

so can ONLY form 3 covalent bonds…

SOLUTION : change one DOUBLE BOND -> DATIVE BOND!

idea behind covalent bond,  non metals want to get octet by sharing electons (you get one from me, at the same time i get one from you)

but in this case, POOR old N cannot accommodate so many electrons, so instead of sharing electrons, it has no choice but to share 2 electrons in a dative with an O without getting ANY electrons in return

SHAPE TIME

So, we have

O                 O

N

O-

3 bond pairs (dative bond counted as covalent bond -> 1 bond pair)

O lone pair

NO3(-) is trigonal planar = planar

THEN,

ICL4(-)

STEP 1 : I

STEP 2 : assign 1 electron to any CL

STEP 3 :

normal CL need 1 covalent bond

normal CL need 1 covalent bond

normal CL need 1 covalent bond

***CL- no need any covalent bond***

**CL- does not need any electrons from centre atom**

*because all it needs is to be octet*

yet being a part of ICL4, it has to be bonded to the centre atom (STEP 1) somehow… and that is a DATIVE BOND

lols yeah it looks weird but that is how it goes

so we got

CL

CL              I             CL

CL-

STEP 4

I has 7 electrons

after 3 covalent bonds with the CLs, still need add in 4 more electrons

so altogether…

4 bond pairs (each with 1 CL, including the dative bond CL)

2 lone pairs (group the 4 extra I electrons into 2 lone pairs)

so shape is square planar

Square planar = planar

OK! last one is the answer, lets go thr

PO4(3-)

STEP 1 : P

STEP 2 : 1 electron each to 3 lucky Os!

STEP 3 :

normal O needs 2 covalent bonds

O- needs 1 covalent bond

O- needs 1 covalent bond

O- needs 1 covalent bond

STEP 4 :

we have

O-                      O-

P

O-                    O

so P is GRP V, it has 5 electrons

it forms 5 covalent bonds with the Os,

so all settled for its electrons.

it can expand octet so 10 electrons no problem

OK! consider shape

4 bond pairs

0 lone pairs

tetrahedral

tetrahedral -> not planar

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