Chemistry: Post your doubts here!

[dumbledore.]

Which pair of reactions could have the same common intermediate?
W CH3CH2CH3 → intermediate → (CH3)2CHCN
X CH3CH(OH)CH3 → intermediate → (CH3)2C(OH)CN
Y CH3CH=CH2 → intermediate → CH3CH(OH)CH3
Z CH3CO2CH2CH2CH3 → intermediate → CH3CH2CH2Br
A W and X B W and Y C X and Z D Y and Z

Ans is B

 

[Holmes]

ORGANIC SHORT NOTES
Alkanes
Undergoes free radical substitution
Ex - Ethane + Cl2
Conditions : UV light
Ex -
Initiation Cl2 --> 2Cl*
Propagation steps : (a) C2H6 + Cl* --> C2H5* + HCl (b) C2H5* + Cl2 --> C2H5Cl + Cl*
Termination steps : (a) C2H5* + Cl* --> C2H5Cl (b) Cl* + Cl* --> Cl2 (c) C2H5* + C2H5* --> C4H10

Alkenes
Undergoes electrophyllic addition
(a) With Hydrogen
Conditions : Nickel catalyst , 333K temperature
(b) With water
Conditions : H3PO4 catalyst,water, (273+330)K temperature
(c) With Hot concerntrared KMnO4
The >C=C< bond breaks. If one of the C in the bond have two Carbons attatched to it, it becomes a keytone. Otherwise, it will become a carboxyllic acid.
(d) With cold dilute KMnO4
The >C=C< bond breaks to form a diol. For an example CH2=CH2 will become CH2(OH)CH2(OH).
(e) With Br2 in CCl4
The double bond breaks and the two Bromines go and bond with the two carbons ex - C2H4 --> CH2BrCH2Br
(f) With Br2(aq)
The double bond breaks and one bromine atom bonds with one carbon atom while an OH group go and bond with the other. ex - C2H4 --> CH2BrCH2(OH)
(g) Electrophyllic addition of alkenes
Things to remember : This is a two step reaction.
The Markonikov addition principle : It states that the halogen should go and bond itself with the carbon containing more alkyl (CH3) groups whilst the hydrogen should go and bond with the carbon containg more H atoms. This principle is for HX (where X is a halogen) reaction with alkenes.
(h) Polymerization
Things to remember : Two types, addition polymerization and condensation polymerization
Addition polymerization is with alkenes.
Conditions : High temperature , Highh pressure or High temperature and Zigger-Nata catalyst

Alcohols
Things to remember : Tertiary alcohols cannot be oxidised.
(a) Dehydration
Add conc H2SO4 and heat gently. An alkene will form.
(b) Reaction with PCl5,PCl3 and SOCl2
Replaces OH with a Cl atom.
For PCl3 and SOCl2, it is major condition to heat but not with PCl5.
PCl5 + CH3OH --> CH3Cl + POCl3 + HCl
SOCl2 + 2CH3OH --> SO2 + CH3Cl + H2O
(c) Oxidation
Secondary alcohols become a keytone.
Primary alcohols become carboxyllic acid if heated under reflux. Becomes and aldehyde if it is heated and distilled off.
(d) With Na
2CH3OH + 2Na --> 2CH3ONa +H2
Effervesence etc is observed

Carboxyllic acids and esters
Carboxyllic acids react with NaOH, Na, Na2CO3 etc. Also it reacts with alcohols to make esters. Another thing is that it is partially soluble in water.
(a) Reaction with NaOH, Na and Na2CO3
CH3CO2H + Na --> CH3CO2Na + 1/2H2
CH3CO2H + NaOH --> CH3CO2Na + H2O
2CH3CO2H + Na2CO3 --> 2CH3CO2Na + CO2 + H2O
(b) esterification
Conditions : Add conc H2SO4 and the alcohol. Heat gently.
The C-OH bond in the carboxyllic acid breaks to bond with the alcohol.
CH3CO2H + CH3OH --> CH3CO2CH3 + H2O

ESTERS
(a) Reaction with acids and alkalines
i) acid --> CH3CO2CH3
The ester bond breaks to form the initial carboxyllic acid and alcohol.
ii) alkaline --> CH3CO2CH3
The ester bond breaks to form sodium--thoxide + alcohol
ex -
FOR ACIDS
CH3CO2H --> CH3CO2H + CH3OH
FOR ALKALINE
CH3CO2H ---> CH3CO2Na + CH3OH

Good work boy !

 

[Mstudent]

Which pair of reactions could have the same common intermediate?
W CH3CH2CH3 → intermediate → (CH3)2CHCN
X CH3CH(OH)CH3 → intermediate → (CH3)2C(OH)CN
Y CH3CH=CH2 → intermediate → CH3CH(OH)CH3
Z CH3CO2CH2CH2CH3 → intermediate → CH3CH2CH2Br
A W and X B W and Y C X and Z D Y and Z

Ans is B

The intermediate for
W: CH3CHCH3+ (Hydrogen is lost from central carbon )
X: CH3C(OH)CH3+ (Hydrogen is lost from central carbon)
Y:CH3CHCH3+ (Hydrogen is ADDED to Ch2 due to markonikov's rule)
Z:CH3CH2CH2OH (Alcohol from ester hydrolysis)

So W and Y have the same intermediates!

hope this helped

 

[dumbledore.]

Ans: C

 

[Zaki ali asghar]

Yo, listen, you confused it with hydrolysis of ester with dilute Sodium Hydroxide which does result in a sodium salt of carboxyllic acid and the alcohol!

thanks man,got it.

 

[dumbledore.]

Ans: C

1. Neutral.. Carboxyl group acidic iss liye D k ilawa sab neutral
2. Orange ppt with 2 4 dnph...SIRF aldehyde aur ketones dete hain ppt. B mein ester hai so NO
3. HCl with PCl5...sirf alcohol aur carboxylic acid dete hain. Iss liye C

 

[Mstudent]

1. Neutral.. Carboxyl group acidic iss liye D k ilawa sab neutral
2. Orange ppt with 2 4 dnph...SIRF aldehyde aur ketones dete hain ppt. B mein ester hai so NO
3. HCl with PCl5...sirf alcohol aur carboxylic acid dete hain. Iss liye C

Could you plz translate it in english so that I can understand (I don't understand urdu very well even tho Im Bangladeshi!)

 

[dumbledore.]

Could you plz translate it in english so that I can understand (I don't understand urdu very well even tho Im Bangladeshi!)

1. It says it is neutral. So D option is incorrect as carboxyl group is present.
2. Orange ppt can only be given by aldehydes and ketones. So B option is incorrect as it contain ester group.
3. Only alcohol and carboxylic groups gives HCl with PCl5.
Hence C is the right answer.

 

[Mstudent]

1. It says it is neutral. So D option is incorrect as carboxyl group is present.
2. Orange ppt can only be given by aldehydes and ketones. So B option is incorrect as it contain ester group.
3. Only alcohol and carboxylic groups gives HCl with PCl5.
Hence C is the right answer.

OH ok, thank you very much!

 

[darks]

Both of these please. Thanks : )

 

[DeadbeatCIE]

I have no idea about the kind of approach towards these type of questions any help would be kindly appreciated.

S16qp11

 

[Thisansa]

View attachment 62398 View attachment 62399
I have no idea about the kind of approach towards these type of questions any help would be kindly appreciated.

S16qp11

For question number 3:
First you have to find the Mr of the whole molecule= 207.2+ 4[(12*2)+(1*5)] = 323.2
Then you gotta find the total mass of carbon= 4(12*2)= 96
Then to find the percentage of Carbon in the molecule you have to divide the total mass of C by the total mass of the molecule and multiply by 100.
% of C = (96/323.2)*100 = 29.7%
so the answer is C

For question number 17:
You have to build up an equation for the neutralization reaction taking place which will be CaCO3 +2HCl --> CaCl2 + CO2 + H2O
Then find the number of mol of HCl reacted by equation n=cv = 0.5*(36/1000)=0.018mol
The amount of mol of CaCO3 reacted will be half of that because the molar ratio is 1:2 so, mol of CaCO3= 0.018/2 = 0.009mol
Next we have to find the Mr of CaCO3 which is = 40.1 + 12 + (16*4) = 100.1
then we can find the mass of CaCO3 by rearranging the equation n=m/Mr to m= n*Mr
So mass of CaCO3 will be = 0.009*100.1= 0.9 g
Then the percentage of CaCO3 in the rock can be found by dividing mass of CaCO3 by mass of rock and multiplying by 100
So, %of CaCO3= (0.9/2)*100 = 45%
Thus, B is the answer.

Hope it helps!

 

[Thisansa]

View attachment 62394 View attachment 62395
Both of these please. Thanks : )

For question 4:
The equation we must use is pV=nRT, rearrange the equation to get p=nRT/V
since n and R are constant you can consider that pressure is directly proportional to T and is inversely proportional to V
so you can build up 2 equations p1/p2 = T1/T2 and p1/p2 = V2/V1
we can combine these two equations and get p1/p2 = T1/T2 * V2/V1
so then its just substitutions where p1= 1*10^5
T1= (273+20)= 293 K
T2= (273+100)= 373K
V1= 1V
V2= 1V+ 3V= 4V
If you substitute correctly you'll get 3.18*10^4 so the answer is A

 

[amina1300]

Q 11 20 29
http://papers.gceguide.com/A Levels/Chemistry (9701)/9701_w16_qp_13.pdf

 

[amina1300]

http://papers.gceguide.com/A Levels/Chemistry (9701)/9701_w16_qp_12.pdf
Q30 Q26

 

[Mstudent]

Q 11 20 29
http://papers.gceguide.com/A Levels/Chemistry (9701)/9701_w16_qp_13.pdf

29 should be A. count the number of peaks before the =O, the peaks represent the number of carbon atoms

 

[Metanoia]

Cyanohydrins can be made from carbonyl compounds by generating CN–
ions from HCN in the
presence of a weak base.
In a similar reaction, –
CH2CO2CH3 ions are generated from CH3CO2CH3 by strong bases.

Which compound can be made from an aldehyde and CH3CO2CH3 in the presence of a strong
base?
A CH3CH(OH)CO2CH3
B CH3CO2CH2CH(OH)CH3
C CH3CH2CH(OH)CH2CO2CH3
D (CH3)2C(OH)CH2CO2CH3

this question is from w09qp1. pls help. the ans is C but I dont know why

Hopefully, the explanation in this video would help

 

[Metanoia]

Any trick for this kind of question? :/

Acids can be viewed as proton donors, for options A, B and D, the underlined substances are acids as they donated (lost their protons)

Option B, on the other hand, lost OH-.

 

[Rohith99]

Hey I need help in the Mcq paper 11 for chemistry which I'm gonna write on Wednesday 7th June if anyone writes it before 5am gmt pls share the question paper or the questions which come in it its a deep request someone pls help..it would be a huge favor for me.. Thanks.. Reply asap of anyone sees this msg

 

[Metanoia]

CAN SOMEONE THIS QUESTION??
View attachment 62348View attachment 62349

ANSWER = B