I am in the process of reading through the above mentioned doc from here.

In page-21, second paragraph in section 8.5 I/O Memory, line 6 is given as below:


  When using the I/O specific commands IN and OUT, the I/O addresses 0x00 - 0x3F must be used.


However, in page-224, row 1 of table, first column, Assembly Code Example has the last line as follows:

out TWCR, r16


But as per page-21 statment, above instruction is wrong as TWCR has memory address 0xBC as given in page-396 row 6. 

Is there something wrong here, or am I understanding it wrong?

Please help.

Edit: definitions file


Question: Why does cyclopropane give bromine water test?
Answer: The following ring opening reaction will occour:





You are quite right about the angle strain. Because orbital interactions are not optimal in this geometry. Consider $p$-orbitals, then a natural bond angle would be $\theta\in [90^\circ; 180^\circ]$. A mixing of $s$ and $p$ type orbitals allows a wide range of angles $\theta\in (90^\circ,\dots, 180^\circ)$.

In the cyclo propane $\ce{C3H6}$ - which you can also describe as trimethylene $\ce{(CH2)3}$ -   bonds have to be bent to overlap at all. A possible way of describing the bonding situation is regarding each $\ce{CH2}$ entity as $sp^2$ hybridised. Two of these orbitals are used for $\ce{C-H}$-bonds (not shown) and one forms an inner two-electron-three-centre $\sigma$ bond (left). Leaving $p$-orbitals to form some kind of degenerate $\pi$-like orbitals (middle, right). 



This very general approach can be derived from a Walsh diagram. Schwarz et.al. {@academia.edu} and Hoffmann {@roaldhoffmann.com} described bonding quite similar and it is in quite good agreement with a calculation (BP86/cc-PVTZ, $D_{3h}$) I have done. From this I have prepared a chart of all occupied molecular orbitals formed from valence orbitals and the LUMO. Preview:



Especially the symmetrical orbital 8 resembles very well the schematics. A quite rigorous approach for this theory can also be found here.

It is noteworthy - as mentioned by ron - that there is no notable increase in electron density in the centre of the ring. This may be due to the fact that there are much more orbitals having nodes in the centre than there are without.



Now bromine is known to be easily polarised $\ce{{}^{\delta+}Br-Br^{\delta-}}$ and may intercept at any point of the ring causing a bond break and relaxation to a less strained structure. It will most likely attack at the the $\pi$ type orbitals since bromine is an electrophile. The mechanism is analogous to the addition of bromine to ethene, which is nicely described at chemguide.co.uk. The essential part is the attack of the bromine at the HOMO(s).



The ring opening reaction can be reversed by adding sodium.

However, when there are bromine radicals present (UV light) then substitution will occur:
\begin{aligned}\ce{
Br2 &amp;-&gt;[\ce{h\nu}] 2Br.\\
&amp;+(CH2)3 -&gt; (CH2)2(CHBr) + HBr
}\end{aligned}