oxidation of alkanol

uncle michael, 唔係想挑你機,不過恐怕你解釋唔到
點解唔加大量dilute sulphuric acid而要加小量既conc. sulphuric acid
以下係我既愚見:
oxidation stage:
alkanol (or alcohol) --> alkanal (or aldehyde) --> alkanoic (carboxylic) acid
個人認為sulphuric acid,作為一個catalyst,
去acidified 個 alkanal,令alkanal變成alkanoic acid
冇sulphuric acid, 個 reaction 到左 alkanal就會停左
sulphuric acid愈conc, acidify既過程愈有效,愈會變成alkanoic acid
(A level既解釋就係,愈conc 既sulphuric acid, push the acidifying equilibrium right further)

多多指教

2008-11-18 10:26:22 補充：
其實呢個係YEAR 2既lecture notes黎,唔係我亂UP..

2008-11-18 11:33:52 補充：
sulphuric acid 絕對係catalyst, 過程中會regenerate
而呢個唔係單純睇兩條equation就可以睇到既

full mechanism: (alkanol --> alkanal)
H2SO4 + 2H2O --> 2H3O(+) + SO4(2-)
Cr2O7(2-) + 2H3O(+) = 2CrO3 + 3H2O (加入左兩個H3O+(from sulphuric acid))

2008-11-18 11:34:16 補充：
R-CH2-OH(alkanol) + CrO3 --> R-CH2-O(+)HCrO3(-) (R=carbon chain)
H2O + R-CH2-O(+)HCrO3(-) --> H3O(+) + R-CH2-OH-CrO3(-)
(第一個H3O+ regenerated)
H2O + R-CH2-OH-CrO3(-) --> H3O(+) + R-CHO(alkanal) + HCrO3(-)
(第二個H3O+ regenerated)

2008-11-18 11:34:50 補充：
(alkanal --> alkanoic acid)
H2SO4 + 2H2O --> 2H3O(+) + SO4(2-)
Cr2O7(2-) + 2H3O(+) = 2CrO3 + 3H2O (再加入兩個H3O+(from sulphuric acid))

2008-11-18 11:35:17 補充：
R-CHO + H2O --> R-CH(OH)2
R-CH(OH)2 + CrO3 --> R-CH(OH)-O(+)H-CrO3(-)
R-CH(OH)-O(+)H-CrO3(-) + H2O --> R-CH(OH)-O-CrO3(-) + H3O(+)
(第一個H3O+ regenerated)
R-CH(OH)-O-CrO3(-) + H2O --> R-COOH(alkanoic acid) + H3O(+) + CrO4(2-)
(第二個H3O+ regenerated)

2008-11-18 11:35:35 補充：
有一點必須say sorry, 既然係catalyst, 好似uncle Michael所講,係唔會影響個equilibrium position 既

2008-11-18 11:49:30 補充：
如果我地淨係想要aldehyde既話,
加H2SO4, CrO4(2-)然後distill既話,個yield會好低,
因為當中有一部分既alkanol會react變左alkanoic acid

要提高個yield, 咁就要喺alkaline 既condition 進行
eg1(PCC): Cr2O7(2-), pyridine (a base)
eg2(Swern Oxidation): dimethyl sulfoxide(CH3S=OCH3), (COCl)2, (CH3CH2)3N (a base)

2008-11-18 11:51:35 補充：
呢兩個方法淨係出到alkanal, 出唔到carboxylic acid
咁個yield會高好多(>90%)

2008-11-18 12:05:38 補充：
而家啲synthetic chemist 多數用acidic condition 既Oxidation黎synthesize carboxylic acid
而唔會整aldehyde, aldehyde喺acidic condition非常唔stable,會好快變左acid,
好難control到去到aldehyde就即刻比你distill到出黎
而且,單單用distillation好難refine出pure既product,

2008-11-18 22:20:59 補充：
sorry,我都覺得自己為argue而argue了...

Uncle Michael 都唔係第一次講錯野啦
佢有一次講錯左都俾我ban左
講返真係唔好意思喇....

2008-11-18 21:14:34 補充：
兩位都長篇大論
直接講conc. sulphuric acid of few drops is used to act as catalyst to complete the oxidation of alkanol.
再多的什麼大學notes, 再深入的解釋, 請視乎發問者的理解能力和他所學的層次, 才發表吧

In performing the oxidation of alkanol, a few cm3 of the alkanol and a few cm3 of concentrated potassium dichromate solution are mixed, and then a few drops of concentrated sulphuric acid is added.

In other words, only a small volume of concentrated sulphuric acid is added to the reaction mixture. The advantages are:

(1) The concentration of the other reactants would not be significantly diluted as the volume of concentrated sulphuric acid would not affect much the total volume.

(2) The concentrated sulphuric acid is dilute to a suitable level as the volume of the reaction mixture is much greater than the volume of concentrated sulphuric acid. Therefore, in the reaction mixture, the sulphuric acid becomes suitably diluted.
=

2008-11-18 08:52:52 補充：
To: violinist

(1) If add a large volume of dilute sulphuric acid is used instead, all other reactants would be significantly diluted, followed by a lowered reaction rate. This is not favourable.

2008-11-18 08:53:38 補充：
(2) In this reaction, sulphuric acid acts as a reactant, but not a catalyst as you thought. The definition of a catalyst is “a substance which can alter the reaction rate but is chemically unchanged at the end of the reaction.” Refer to the equations of the reactions:

2008-11-18 08:55:11 補充：
K2Cr2O7 + 4H2SO4 + 3RCH2OH ® K2SO4 + Cr2(SO4)3 + RCHO + 7H2O
K2Cr2O7 + 4H2SO4 + 3RCHO ® K2SO4 + Cr2(SO4)3 + RCOOH

2008-11-18 08:55:26 補充：
Obviously, sulphuric acid is consumed in the reaction, and thus it is NOT a catalyst. Moreover, according the above two equations, sulphuric acid is needed in the formation of both the aldehyde (alkanal) and the carboxylic acid (alkanoic acid).

2008-11-18 08:56:36 補充：
(3) Whether the product is aldehyde or carboxylic acid is NOT determined by the concentration of sulphuric acid. In practice, to get the aldehyde, a simple distillation is performed during the preparation. To get the carboxylic acid, the reaction mixture is heated under reflux.

2008-11-18 08:57:00 補充：
(4) The reactions are reversible. So, think about the rates of the reactions but NOT the shifts in the equilibrium positions.

2008-11-18 08:59:33 補充：
Moreover, it is not desirable to put too much personal idea (個人認為) on the facts known in science.

2008-11-18 15:10:31 補充：
純粹討論，若有言語誤會，請勿見怪。

(5) 單從 equation 中 H2SO4 耗用了，便可以證明它不是 catalyst，因為不符合 定義中的 chemically unchanged at the end of the reaction。

2008-11-18 15:11:02 補充：
(6) 對所提出的 mechanism 有一個很大的疑問，因為只見 alkanol 分階段氧化，但卻沒有 oxidizing agent。可能這是不完全的，沒有了耗用 H2SO4 變成 Cr^3+ 的一部份。

2008-11-18 15:13:03 補充：
(7) 用我描述的方法去製造 aldehyde，不是好方法，通常只局限於細小的分子才可以，但預科程度是可以的。我代課教中七時，通常教以 Ag 作 catalyst，把 alkanol 脫水製成 alkanal。

2008-11-18 15:14:24 補充：
(8) 只有 reversible reaction 才可以有 equilibrium。但在反應條件下，沒法進行 backward reaction（把 Cr^3+ 變回 Cr2O7^2-），所以是 irreversible reaction，沒法談 equilibrium。

2008-11-18 15:15:49 補充：
討論已離原題目很遠了。若想繼續討論，歡迎電郵給我。

2008-11-19 17:17:44 補充：
To 冷風：

首先謝謝你的意見。

當你看清楚題目的時候，題目是問為甚麼要用濃的硫酸。我的回答已相當清楚，是因為小量的濃硫酸不會稀釋其他反應物，而濃硫酸卻會被反應混合物稀釋至適當濃度。

我的答案是在正面的寫法，但其實已包含 violinist 為甚麼不用大量稀硫酸（其實即是我答案的反面，會把其他反應物稀釋）。但 violinist 還提出一些意見，我才寫出一些回應，但已註明是給 violinist 的，而不是給問題者補充答案，因為答案已非常清晰地回答了問題。

2008-11-19 17:18:05 補充：
在整個反應中，硫酸並非催化劑。而且無論是否催化劑，都不是問題的答案。

我常打錯字，又常匆匆地作答，犯小錯誤，但大錯總是很少。而且已忘記了我錯些甚麼，令你有這麼大的權力令你可以在這裡「BAN」了發問者或回答者。

真的，用心回答不犯錯（無論大、小錯）很難，去挑掦別人犯錯很容易。但我卻捨易取難。