• We need your support!

    We are currently struggling to cover the operational costs of Xtremepapers, as a result we might have to shut this website down. Please donate if we have helped you and help make a difference in other students' lives!
    Click here to Donate Now (View Announcement)

Chemistry: Post your doubts here!

Messages
537
Reaction score
358
Points
73
Should be B. The chiral carbon is the rightmost carbon on that molecule. And when viewed individually, none of the C=C carbons are holding two identical groups. ( although this seems to be more correctly known as E/Z isomers rather than cis trans)
i thought the same but there aren't any similar groups or elements present on the opposite sides of C atoms representing CIS or trans
 
Messages
7
Reaction score
4
Points
13
please help me with question no. 2
 

Attachments

  • upload_2015-4-3_13-8-15.png
    upload_2015-4-3_13-8-15.png
    314.1 KB · Views: 7
Messages
603
Reaction score
1,102
Points
153
please help me with question no. 2
Based on 1st equation, 1 mole of NaN3 produces 1 mole of Na and 1.5 moles of N2.

Based on the 2nd equation, the 1 mole of Na produced in the 1st reaction will then produce another 0.1 mol of N2.
 
Messages
204
Reaction score
565
Points
103
Can anyone help me that how we can distinguish between primary secondary and tertiary helogenoalkens? ??
And secondly if anyone is having the organic notes for AS chem plz share them with me. .I desperately need them.
 
Messages
603
Reaction score
1,102
Points
153
Can anyone help me that how we can distinguish between primary secondary and tertiary helogenoalkens? ??
And secondly if anyone is having the organic notes for AS chem plz share them with me. .I desperately need them.

Theoretically, one way would be to hydrolyze them with NaOH to form primary secondary and tertiary alcohols and then attempt to oxidize them with KMno4

1) tertiary would not be oxidized (no decolorization of KMnO4)
2) Primary would be oxidized to acid, secondary would be oxidized to ketone. Test with 2,4-DNPH for the ketone.
 
Last edited:
Messages
204
Reaction score
565
Points
103
Theoretically, one way would be to hydrolyze them with NaOH to form primary secondary and tertiary alcohols and then attempt to oxidize them with KMno4

1) tertiary would not be oxidized (no colorization of KMnO4)
2) Primary would be oxidized to acid, secondary would be oxidized to ketone. Test with 2,4-DNPH for the ketone.
And can u plzzzzz share that dnph equation. .???
 
Messages
1,318
Reaction score
1,374
Points
173
Capture.JPG
Check out the attached question
The answer is A (1,2 and 3 are correct)
But V3+ means that it has 20 electrons
so that means it has a configuration of 1s2 2s2 2p6 3s2 3p6 4s2
this shows that all the orbitals are filled right?
Then how come it still has unpaired electrons?
 
Messages
1,318
Reaction score
1,374
Points
173
Why is it D... I thought it would be C because
- pale yellow ppt is formed by Br- when reacted with Silver nitrate
- white ppt for Cl-
- cream ppt for I-
and
- Cl- is soluble
- I- is soluble in concentrated ammonia
- Br- is insoluble

So if it is partly dissolved, and a darker yellow ppt is present, shouldn't Br- be present instead of I- because I- will also dissolve therefore there will be no ppt remaining after ammonia is added.....???
 

Attachments

  • Capture.JPG
    Capture.JPG
    45.2 KB · Views: 4
Messages
1,318
Reaction score
1,374
Points
173
Capture.JPG
Shouldn't the electronic configuration be
1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2
Since the 4s orbital is filled before the 3d orbital
But the mark scheme says
1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 5s2
But i have been taught that 4s fills before 3d so which one is correct??
 
Top