Why can't I send this IP packet?

I'm trying to send an IP packet using c#.

    destAddress = IPAddress.Parse("192.168.0.198"),
    destPort = 80;

    // Create a raw socket to send this packet
    rawSocket = new Socket(AddressFamily.InterNetwork, SocketType.Raw, ProtocolType.IP);

    // Bind the socket to the interface specified
    IPEndPoint iep = new IPEndPoint(IPAddress.Parse("192.168.0.140"),0);
    rawSocket.Bind(iep);

    // Set the HeaderIncluded option since we include the IP header
    rawSocket.SetSocketOption( socketLevel, SocketOptionName.HeaderIncluded, 1 );

    // Send the packet!
    int rc = rawSocket.SendTo(builtPacket, new IPEndPoint(destAddress, destPort));
    Console.WriteLine("sent {0} bytes to {1}", rc, destAddress.ToString());

The content of builtPacket is shown below. It's an IP packet containing a TCP SYN packet (That's what I think I created anyway).

45 00 00 28 00 00 00 00 02 06 36 6E C0 A8 00 8C

C0 A8 00 C6 14 1E 00 50 00 00 00 00 00 00 00 00

05 02 FF FF E6 4F 00 00

The output is:

sent 40 bytes to 192.168.0.198

The problem is I don't see anything in the Wireshark trace. It's like the data is not getting far enough down the stack for Wireshark to see it? If I use a browser to connect to 192.168.0.198, Wireshark shows all the packets, but shows nothing when I try to send a packet using the above code and data.

My config:

I am running as admin so it's not a permissions problem.
Windows7 ( Not running in a VM)
Wireless connection only (IP config reports its IP as 192.168.0.140)

What am I doing wrong?

I'm sure Occam's Razor applies here, but I've been looking at this for hours and can't figure out what's wrong.

回答1:

This question, backed up by MSDN, claims that Windows no longer (XP SP 2 through 7) allows transmission of TCP data using raw sockets.

回答2:

My guess is that either Wireshark is not looking at the right network interface, or that the destination ip address somehow resolves to the local machine, in which case it will routed inside of the OS and be invisible to the 'Shark.